# Statistics 1

Relationship between A.M. and G.M.
As we have seen above the formula for the Arithmetic mean and the Geometric mean are as follows:

where a and b are the two given positive numbers.

Relationship between A.M. and G.M.

Let A and G be A.M. and G.M.

So,

Now let’s subtract the two of them

This shows that A ≥ G

Example

Find the two numbers, If Arithmetic mean and Geometric mean of two positive real numbers are 20 and 16, respectively.

Solution:

Given

Now we will put these values of a and b in

(a – b)2 = (a + b)2 – 4ab

(a – b)2 = (40)2 – 4(256)

= 1600 – 1024

= 576

a – b = ± 24 ( by taking the square root) …(3)

By solving (1) and (3), we get

a + b = 40

a – b = 24

a = 8, b = 32 or a = 32, b = 8

Properties of relationship of A.M and G.M.
Properties of relationship of A.M and G.M.

Property I: If the Arithmetic Mean and Geometric Mean of two positive numbers a and b are A and G respectively, then

A > G

As A – G > 0

Proof:

Let A and G be the Arithmetic Means and Geometric Means respectively of two positive numbers a and b.

Then,

As, a and b are positive numbers, it is obvious that A > G when G = -√ab.

Now we have to show that A ≥ G when G = +√ab.

We have,

Therefore, A – G ≥ 0 or, A ≥ G.

This proves that the Arithmetic Mean of two positive numbers can never be less than their Geometric Means.

Property II: If A be the Arithmetic Mean and G be the Geometric Mean between two positive numbers a and b, then the quadratic equation whose roots are a, b is

x2 – 2Ax + G2 = 0.

Proof:

As, A and G are the Arithmetic Mean and Geometric Mean respectively of two positive numbers a and b then, we have

The equation having a, b as its roots is

Property III: If A is the Arithmetic Means and G be the Geometric Means between two positive numbers, and then the numbers are A ± √A2 – G2

Proof:

Since, A and G be the Arithmetic Means and Geometric Means respectively then, the equation having its roots as the given numbers is

Example

Find two positive numbers whose Arithmetic Means increased by 2 than Geometric Means and their difference is 12.

Solution:

Let the two numbers be a and b. Then,

a-b = 12 …………………… (i)

Given

Now, a – b = 12

Solving (ii) and (iii), we get a = 16, b = 4

Hence, the required numbers are 16 and 4.

Example

If 34 and 16 are the Arithmetic Means and Geometric Means of two positive numbers respectively. Find the numbers.

Solution:

Let the two given positive numbers are a and b.

Given

Arithmetic Mean = 34

Geometric Mean = 16

As we know that

(a-b)2 = (a+b)2 – 4ab

⇒ (a-b)2 = (68)2 – 4 × 256 = 3600 ( from equation (i) and (ii))

⇒ a-b = 60…………………………. (iii)

After solving (i) and (iii), we get the numbers a = 64 and b = 4.

Hence, the required numbers are 64 and 4.